Simplify; express your answer in exponential form. Assume $q\neq 0, k\neq 0$. $\dfrac{{(q^{-4}k)^{-4}}}{{(q^{-2}k^{-1})^{-4}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(q^{-4}k)^{-4} = (q^{-4})^{-4}(k)^{-4}}$ On the left, we have ${q^{-4}}$ to the exponent ${-4}$ . Now ${-4 \times -4 = 16}$ , so ${(q^{-4})^{-4} = q^{16}}$ Apply the ideas above to simplify the equation. $\dfrac{{(q^{-4}k)^{-4}}}{{(q^{-2}k^{-1})^{-4}}} = \dfrac{{q^{16}k^{-4}}}{{q^{8}k^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{16}k^{-4}}}{{q^{8}k^{4}}} = \dfrac{{q^{16}}}{{q^{8}}} \cdot \dfrac{{k^{-4}}}{{k^{4}}} = q^{{16} - {8}} \cdot k^{{-4} - {4}} = q^{8}k^{-8}$